1000\(cm^{3}\) of HClO_{4} [/tex] contain 0.00046mol
16\(cm^{3}\) of HClO_{4} [/tex] contain x
x = \(\frac{16cm ^{3} * 0.00046mol}{1000cm x^{3} }\)
x = 0.00000736mol
\(M_{1} V_{1} = M_{2} V_{2}\)
(0.00000736mol)(16\(cm^{3}\))= \((x) (42cm^{2})\)
\(x= \frac{(0.00000736mol) * (16cm^{3}) }{42cm^{3} }\)
= 0.000002803mol
pH= -log [\(H^{+}\)]
= -log (0.000002803cm\(^{3}\)
= 5.55
